What is the integral from 1 to 4 of ln(sqrt(x)+1) ?

The integral from 1 to 4 of ln(sqrt(x)+1) is -1/2+3ln(3)

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integral from 1 to 4 of ln(sqrt(x)+1)

\int_{1}^{4} ln (x + 1) d x

- \frac{1}{2} + 3 ln (3)

Decimal

2.79583 \dots

Solution steps

\int_{1}^{4} ln (x + 1) d x

Apply u-substitution

: \int_{2}^{3} ln (u) \cdot 2 (u - 1) d u

= \int_{2}^{3} ln (u) \cdot 2 (u - 1) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{2}^{3} ln (u) (u - 1) d u

Apply Integration By Parts

: [ln (u) (\frac{u ^{2}}{2} - u) - \int \frac{1}{2} (u - 2) d u]_{2}^{3}

= 2 [ln (u) (\frac{u ^{2}}{2} - u) - \int \frac{1}{2} (u - 2) d u]_{2}^{3}

\int \frac{1}{2} (u - 2) d u = \frac{u ^{2}}{4} - u

= 2 [ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{2}^{3}

Simplify

2 [ln (u) (\frac{u ^{2}}{2} - u) - (\frac{u ^{2}}{4} - u)]_{2}^{3} : 2 [\frac{1}{2} u^{2} ln (u) - u ln (u) - \frac{u ^{2}}{4} + u]_{2}^{3}

= 2 [\frac{1}{2} u^{2} ln (u) - u ln (u) - \frac{u ^{2}}{4} + u]_{2}^{3}

Compute the boundaries:

- \frac{1}{4} + \frac{3}{2} ln (3)

= 2 (- \frac{1}{4} + \frac{3}{2} ln (3))

Simplify

= - \frac{1}{2} + 3 ln (3)

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What is the integral from 1 to 4 of ln(sqrt(x)+1) ?
The integral from 1 to 4 of ln(sqrt(x)+1) is -1/2+3ln(3)